Practical exercises - Answers
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The procedure we employ in this case is exactly the same as in Exercise
6, with the exception that in the Frequencies:
Statistics dialog box we activate variance,
standard deviation and variance. When
we click OK, we obtain all three measures, which are as shown below.
Descriptive
statistics
|
|
N |
Range |
Std. Deviation |
Variance |
|
Test scores |
48 |
64 |
15.631 |
244.339 |
|
Valid N (listwise) |
48 |
|
|
|
The most important figure here is standard deviation, which indicates
that the average distance of the scores from the mean is 15.631. In simple
terms this suggests that the average dispersion of test scores in this class is
about 15.6 points above and below the mean. (Note that the variance is the
square of standard deviation.) Given that the analysis in the previous two
questions found that the mean of the distribution is 70.54, one can argue that
on average a large part of the scores will be between 70.54 + 15.6 and 70.54 -
15.6, namely between 86 and 55. In addition, the range indicates that the
distance between the highest and the lowest score in the distribution is 64
points. It depends on the teacher's educational philosophy and expectations
whether this is considered to be satisfactory. Be that as it may, the
information provided by these figures is most useful.
Mean, range and standard
deviation.
The procedure employed to compute the mean, range and standard deviation is the
same as that employed in the previous question. The two sets are set in
separate columns as two different distributions of examination scores and
within a separate variable (for example, 'psychol' and 'sociol' respectively).
The computation produces the following findings for the performance of the two
classes:
Descriptive
Statistics
Psychology
|
|
N |
Range |
Mean |
Std. Deviation |
Variance |
|
psychol |
15 |
5 |
6.13 |
1.356 |
1.838 |
|
Valid N (listwise) |
15 |
|
|
|
|
Descriptive
Statistics
Sociology
|
|
N |
Range |
Mean |
Std. Deviation |
Variance |
|
sociol |
15 |
6 |
6.33 |
1.718 |
2.952 |
|
Valid N (listwise) |
15 |
|
|
|
|
This shows that overall the students performed slightly better in
sociology than in psychology, however, the scores in sociology were spread more
than in psychology. This is evident in both, the range and standard deviation.
Whether these differences are significant enough to say that overall students
did better in sociology than in psychology is not shown in these results. For
this to be determined we must conduct a significance test.
z-scores. The computation of
z-scores follows a similar path. We compute z-scores separately for each
variable by going to Analyze/Descriptive
Statistics/Descriptives., and then transferring the variable in question to
the Variable(s) box. Then activate Save standardized values as variables,
and finally click OK.
As soon as the computation is completed, the z-scores for each score of
the distribution is added to the set and is visible in the data editor:
The computation here is simple but cumbersome simply because of the
nature of the data. As you see, they are presented in a table. We therefore
have to convert the table in a manner that would make the data entry easier.
Converting the table
During this
step we translate the cells of the table into numbers indicating the place of
each cell in terms of its position within the table. Following this, cell 612
will be described as 1, 1, 612 (for it is in the first row, it is the first
column, and its value is 612); similarly, cell 220 will be described as 1, 2,
220, meaning it is in the first row, in the second column and its value is 220.
(Note that we discard the totals; they will be calculated by the computer!)
When this description is complete, the table will be as follows:
Converted table
|
Sex |
Tert.
Education |
Count |
|
1 |
1 |
612 |
|
1 |
2 |
220 |
|
2 |
1 |
138 |
|
2 |
2 |
530 |
This data set is ready to be entered in the computer, as shown
here, in the corresponding columns of the Data Editor.
The
definition of the variables is the same as that employed with listed data. The
variables are: sex, tertiary education, and count. Define the variables, as
usual ('gender', 1 for M; 2 for FM; 'tereduc', 1 for Yes, and 2 for No). Set
'count' in the third line.
Entering
the data and processing
Entering the data from a table, particularly with large numbers can be a
very cumbersome and time-consuming task. Here you will be introduced to a
simple method that can reduce this process to a minimum. Here is how we go
about it:
Go to Analyze/Descriptive Statistics/Crosstabs.
In the Crosstabs
dialog box transfer Gender to the
first box under Row(s).
Transfer Tertiary
education to the second box.
Click Data in the
menus and then click Weight cases.
Click the button before Weigh
cases by.
Transfer Count to
Frequency Variable.
Click OK. Then
click Statistics and activate Phi and Cramer's V.
Click OK.
The results are displayed in a table in the Viewer window, and are as
shown below.
Symmetric
measures
|
|
Value |
Approx. Sig. |
|
|
Nominal by
Nominal |
Phi |
.283 |
.000 |
|
|
Cramer's V |
.283 |
.000 |
|
N of Valid Cases |
1900 |
|
|
The Phi shows that the differences between males and females regarding
access to tertiary education are significant (sig.: .000). One can argue that
there are more males with tertiary education than females.
The test that is appropriate to process ordinal data is Spearman's rho.
To conduct this test we begin with defining the variables and then proceed to
enter the data. Here is how we go about addressing this question:
Define the variables, one being husband's ranking and the
other wife's ranking.
Enter the data in the computer as shown above each set in a
column, that is, one in the husband's column and the other in the wife's column.
In the Data Editor's menus click Analyze/Correlation/Bivariate.
In the Correlation
dialog box, transfer both variables to the Variables
box.
Click Spearman's rho and
then OK.
This way you obtain the
following result:
Correlations
|
|
|
|
Husband's ranking |
Wife's ranking |
|
Spearman's rho |
Husband's
ranking |
Correlation
Coefficient |
1.000 |
.927(**) |
|
|
|
Sig. (2-tailed) |
. |
.000 |
|
|
|
N |
10 |
10 |
|
|
Wife's ranking |
Correlation
Coefficient |
.927(**) |
1.000 |
|
|
|
Sig. (2-tailed) |
.000 |
. |
|
|
|
N |
10 |
10 |
**
Correlation is significant at the 0.01 level (2-tailed).
We focus here on the strength and direction of correlation. After all,
this is what we want to know, that is, whether the rankings are correlated, and
if so, if the correlation is strong or weak and if it is positive or negative.
The results show that the correlation coefficient is .927, which indicates a
very strong and a positive correlation. This association is also found to be
significant, shown by the .000 level, as well as by the two asterisks next to
the coefficient, as explained at the foot of the table.
We first define the variables: one being hours of study ('hours') and
the other test scores ('scores'). We then enter the data in the computer as in
the previous question, each set in a column: one under 'hours' and the other
under 'scores'.
Getting a scattergram
To construct the
scattergram we proceed as follows.
Click Graphs in the menus and then click Scatter.
In the Scatterplot dialog box click Simple and then Define.
Transfer Hours to Y Axis and Test scores to X Axis.
Click OK.
The scattergram is displayed in the Viewer window as shown below.
This graph demonstrates clearly that the scores do not indicate any
particular direction, and there is certainly no distinct indication that a
considerable association exists. At best, this can only be a sign of a very low
correlation.
Getting Pearson's r
In the Data Editor's menus click Analyze/Correlation/Bivariate.
In the Correlation
dialog box transfer both variables to the Variables
box and click Pearson's rho and OK.
The Viewer's window shows the following table:
Correlations
|
|
|
Hours of study |
Test scores |
|
Hours of study |
Pearson Correlation |
1 |
.133 |
| Sig. (2-tailed) |
. |
.635 |
|
| N |
15 |
15 |
|
|
Test scores |
Pearson Correlation |
.133 |
1 |
| Sig. (2-tailed) |
.635 |
. |
|
| N |
15 |
15 |
These data show a very low coefficient (.133) and hence a very low correlation between the
variables. A similar direction in the association between the variables is
conveyed by the scattergram.
The process of addressing these questions is identical to that employed
in the previous question. Hence, there is no need to repeat it in this context.
Enter the variables ('accept', and 'satisf') and proceed to the processing. You
should obtain the following:
Scattergram
The scattergram demonstrates clearly a positive and also a strong
correlation. Let us see what Pearson's r reveals.
Pearson's r
Correlations
| Acceptance of power | Satisfaction | ||
| Acceptance of power | Pearson Correlation | 1 | .885(**) |
| Sig. (2-tailed) | . | .001 | |
| N | 10 | 10 | |
| Satisfaction |
Pearson Correlation | .885(**) | 1 |
| Sig. (2-tailed) | .001 | . | |
| N | 10 | 10 |
**
Correlation is significant at the 0.01 level (2-tailed).
Specific answers:
1
The scattergram demonstrates clearly a positive and also a strong correlation.
2
Pearson's r suggests that the correlation between the two variables is positive
and strong.
3
Statistical computation and graphic presentation produce very consistent
results.